If a particle is moving in a straight line such that its distance sin mts at any time tin sec is given by s- t44-2t3-4t2-7- then its acceleration is minimum at which value of -t-

S= t^44 2t^3+4t^2+7 ⇒ v= dsdt=t^3 6t^2+8t ⇒ a=dvdt=3t^2 12t+8 dadt=6t 12 and d^2dt^2=6 gt;0 ∴ 'a' is minimum at dadt =0 ⇒ t=2 ...

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Standard XII

Physics

a-t Graph

If a particle...

Question

If a particle is moving in a straight line such that its distance $s ($ in mts $)$ at any time $t ($ in sec) is given by $s = \frac{t^{4}}{4} - 2 t^{3} + 4 t^{2} + 7,$ then its acceleration is minimum at which value of $^{'} t^{'}$

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Solution

$s = \frac{t^{4}}{4} - 2 t^{3} + 4 t^{2} + 7 \Rightarrow v = \frac{d s}{d t} = t^{3} - 6 t^{2} + 8 t$
$\Rightarrow a = \frac{d v}{d t} = 3 t^{2} - 12 t + 8$
$\frac{d a}{d t} = 6 t - 12$ and $\frac{d^{2}}{d t^{2}} = 6 > 0$
$∴^{'} a^{'}$ is minimum at $\frac{d a}{d t} = 0$
$\Rightarrow t = 2$

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If a particle is moving in a straight line such that its distance s(in mts) at any time t(in sec) is given by s=t44−2t3+4t2+7, then its acceleration is minimum at which value of ′t′

s=t44−2t3+4t2+7⇒v=dsdt=t3−6t2+8t ⇒a=dvdt=3t2−12t+8 dadt=6t−12 and d2dt2=6>0 ∴′a′ is minimum at dadt=0 ⇒t=2

If a particle is moving in a straight line such that its distance $s ($ in mts $)$ at any time $t ($ in sec) is given by $s = \frac{t^{4}}{4} - 2 t^{3} + 4 t^{2} + 7,$ then its acceleration is minimum at which value of $^{'} t^{'}$

$s = \frac{t^{4}}{4} - 2 t^{3} + 4 t^{2} + 7 \Rightarrow v = \frac{d s}{d t} = t^{3} - 6 t^{2} + 8 t$
$\Rightarrow a = \frac{d v}{d t} = 3 t^{2} - 12 t + 8$
$\frac{d a}{d t} = 6 t - 12$ and $\frac{d^{2}}{d t^{2}} = 6 > 0$
$∴^{'} a^{'}$ is minimum at $\frac{d a}{d t} = 0$
$\Rightarrow t = 2$